In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.125 T magnetic field q) 50%

Part (a) What electric field strength, in volts per meter, is needed to select a speed of 41 106 ms?
Part (b) What is the voltage, in kilovolts, between the plates if they are separated by 0.75 cm?

a) E = 5138.25 V/m

b) V = 38.54 V

Explanation:

the following data is available:

B = 0.125 T

v = 41106 m/s

d = 0.75 cm = 0.0075 m

a)

The intensity of the electric field can be calculated based on the following formula:

E = B*v = (0.125 T)*(41106) = 5138.25 V/m

b)

the voltage can be calculated using the formula:

V = E*d = (5138.25 V/m)*(0.0075 m) = 38.54 V

(A) E = 5138V/m

(B) V = 39V

Explanation:

Given B = 0.125T and velocity v = 41106m/s

E = electric field

B = magnetic field

The magnetic field and the electric field are related by the formula E = Bv

The condition for the use of the above formula is that the direction of the velocity, must be perpendicular to the electric and magnetic field. The magnetic and electric l field must also be perpendicular to each other.

Where v = the velocity of the charged particles. The charge move with constant velocity and so these particles do not accelerate, a = 0.

The charged particle moves through the electric and magnetic field that exert equal forces on it that are oppositely directed and as a result cancel out. The electric force is equal in magnitude to Eq while the magnetic force is given by qvB.

By newton's second law

Fm – Fe = 0

qvB – Eq = 0

Eq = qvB

q cancels out on both sides.

E = Bv

E = Bv = 0.125 × 41106 = 5138V/m.

(b) V = Ed

= 5138×0.75×10‐² = 39V

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